Codeforces Beta Round #4 (Div. 2 Only) A. Watermelon【暴力/数学/只有偶数才能分解为两个偶数】

time limit per test

 1 second

memory limit per test

 64 megabytes

input

 standard input

output

 standard output

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed w kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.

Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

Input

The first (and the only) input line contains integer number w (1 ≤ w ≤ 100) — the weight of the watermelon bought by the boys.

Output

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

Sample test(s)

input

8

output

YES

 【分析】：给你一个数，能不能分成两个全是偶数的数。

【分析】：暴力枚举或者由偶数只能->偶数+偶数->该数是除了2以外的偶数就输出YES

【代码】：

#include<bits/stdc++.h>
using namespace std;

#define LL long long
using namespace std;
int n;
int main()
{
    cin>>n;
    int f=0;
    for(int i=1;i<=n/2;i++)
    {
        if((i%2==0)&&((n-i)%2==0))
        {
            f=1;
            break;
        }
    }
    if(f) puts("YES");
    else puts("NO");
    return 0;
}

#include<bits/stdc++.h>
using namespace std;

#define LL long long
using namespace std;
int n;
int main()
{
    cin>>n;
    int f=0;
    if(n%2==0&&n!=2)
        f=1;
    if(f) puts("YES");
    else puts("NO");
    return 0;
}